Question

The population of a town increased to 54000 in 2010 at the rate of 5% per annum. the population in 2008 was​

Answer 1

❥${\green{\underline{\huge{\mathbb{Given:-}}}}}$

•In 2010 , the population of a town=54000

•The population increased at the rate of 5% per annum.

❥${\green{\underline{\huge{\mathbb{To\: Find:-}}}}}$

The population in 2008.

❥${\green{\underline{\huge{\mathbb{Solution:-}}}}}$

Suppose,

The population in 2008= x

Time difference= (2010-2008) years

= 2 years

So,

| Time (n) = 2 years.

| Principal (p) = x

|Rate of interest (r)= 5%

We know,

$Uniform \: rate \: of \: growth \: = p {(1 + \frac{r}{100} )}^{n}$

According to question,

$x {(1 + \frac{5}{100} )}^{2} = 54000 \\ = > x {(1 + \frac{1}{20} )}^{2} = 54000 \\ = > x {( \frac{21}{20} )}^{2} = 54000 \\ = > x \times \frac{21}{20} \times \frac{21}{20} = 54000 \\ = > x = 54000 \times \frac{20}{21} \times \frac{20}{21} \\ = > x = 48979.59(approx)$

❥${\green{\underline{\huge{\mathbb{Answer:-}}}}}$

Hence, the population in 2008 was 48979.59(approx).